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2 years ago

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A motorcycle is travelling at a constant velocity of 30ms. The motor is in high gear and emits a hum of 700Hz. The speed of soun

d in air is 340ms, and there is no wind. a. What is the period of the sound waves emitted from the motorcycle?b. What is the distance travelled by the motorcycle during one period of the sound oscillation?c. What is the distance travelled by a sound wave in air during one period of oscillation?d. The humming of the motorcycle is a sound that is emitted in all directions and, once emitted, it travels in the reference frame of the still air at the speed of sound in air. Consider the emitted sound that is moving directly out in front of the motorcycle

1 answer:

timurjin [86]2 years ago

5 0

Answer:

a) $T=1.43\times 10^{-3}\ s$

b) $d=0.0429\ m$

c) $\lambda=0.4857\ m$

d) $f_o=767.7\ Hz$

Explanation:

Given:

velocity of the sound from the source, $S=340\ m.s^{-1}$

original frequency of sound from the source, $f_s=700\ Hz$

speed of the source, $v_s=30\ m.s^{-1}$

(a)

We know time period is inverse of frequency:

Mathematically:

$T=\frac{1}{f}$

$T=\frac{1}{700}$

$T=1.43\times 10^{-3}\ s$

(b)

Distance travelled by the motorcycle during one period of sound oscillation:

$d=v_s.T$

$d=30\times 1.43\times 10^{-3}$

$d=0.0429\ m$

(c)

The distance travelled by the sound during the period of one oscillation is its wavelength.

$\lambda=\frac{S}{f}$

$\lambda=\frac{340}{700}$

$\lambda=0.4857\ m$

(d)

observer frequency with respect to a stationary observer:

<u>According to the Doppler's effect:</u>

$\frac{f_o}{f_s}= \frac{S+v_o}{S-v_s}$ ...........................(1)

where:

$f_o\ \&\ v_o$ are the observed frequency and the velocity of observer respectively.

Here, observer is stationary.

$\therefore v_o=0\ m.s^{-1}$

Now, putting values in eq. (1)

$\frac{f_o}{700}= \frac{340+0}{340-30}$

$f_o=767.7\ Hz$

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